Question

# What is the molar solubility (s) of ${\mathrm{Ba}}_{3}{\left({\mathrm{PO}}_{4}\right)}_{2}$ in terms of ${\mathrm{K}}_{\mathrm{sp}}$?

Open in App
Solution

## Step 1: Solubility product:${\mathrm{K}}_{\mathrm{sp}}$ can be called a solubility product or the solubility constant.Consider, “the equilibrium between the undissolved solid and the ions in a saturated solution”. This can be represented by the equation: ${\mathrm{Ba}}_{3}{\left({\mathrm{PO}}_{4}\right)}_{2}\left(\mathrm{s}\right)\stackrel{\mathrm{Saturated}\mathrm{solution}\mathrm{in}\mathrm{water}}{⇌}3{\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+{2{\mathrm{PO}}_{4}}^{3-}\left(\mathrm{aq}\right)$The equilibrium constant is given by equation: ${\mathrm{K}}_{\mathrm{sp}}=\frac{{\left[{\mathrm{Ba}}^{2+}\right]}^{3}{\left[{{\mathrm{PO}}_{4}}^{3-}\right]}^{2}}{\left[{\mathrm{Ba}}_{3}{\left({\mathrm{PO}}_{4}\right)}_{2}\right]}$For a pure solid substance, the concentration remains constant and it can be written as: ${\mathrm{K}}_{\mathrm{sp}}=\mathrm{K}\left[{\mathrm{Ba}}_{3}{\left({\mathrm{PO}}_{4}\right)}_{2}\right]={\left[{\mathrm{Ba}}^{2+}\right]}^{3}{\left[{{\mathrm{PO}}_{4}}^{3-}\right]}^{2}$Step 2: Calculation of Solubility product:For solid barium phosphate in equilibrium with its saturated solution, the product of the concentrations of barium and phosphate ions is equal to its solubility product constant. The concentrations of the two ions will be equal to the molar solubility of the barium phosphate. If molar solubility is $\mathrm{S}$, then the solubility product will be: ${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ba}}^{2+}\right]}^{3}{\left[{{\mathrm{PO}}_{4}}^{3-}\right]}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{sp}}={\left[3\mathrm{S}\right]}^{3}{\left[2\mathrm{S}\right]}^{2}$

Suggest Corrections
1
Join BYJU'S Learning Program
Select...
Related Videos
Solubility and Solubility Product
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program
Select...